A 57.5-kg runner has a speed of 5.80 m/s at one instant during a long-distance event.
(a) What is the runner's kinetic energy at this instant?
967.15 J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
im having trouble with this one.
To find the kinetic energy of the runner, you can use the formula:
Kinetic Energy (KE) = (1/2) * mass * speed^2
(a) Substituting the given values, we have:
KE = (1/2) * 57.5 kg * (5.80 m/s)^2
= (1/2) * 57.5 kg * 33.64 m^2/s^2
= 967.1 J (rounded to one decimal place)
Therefore, the runner's kinetic energy at this instant is approximately 967.1 J.
(b) To determine the change in kinetic energy when the runner doubles his speed, we need to compare the initial kinetic energy (KE1) with the final kinetic energy (KE2) after doubling the speed.
Let's assume the initial kinetic energy is KE1 and the final kinetic energy is KE2.
Now, we know that the runner doubles his speed. If the initial speed is v, the final speed is 2v.
Using the formula for kinetic energy, we can write:
KE1 = (1/2) * mass * v^2
KE2 = (1/2) * mass * (2v)^2
Simplifying the equations, we get:
KE1 = (1/2) * mass * v^2
KE2 = (1/2) * mass * 4v^2
= 2 * (1/2) * mass * v^2
= 2 * KE1
Therefore, the final kinetic energy (KE2) is twice the initial kinetic energy (KE1) when the speed is doubled.
In other words, the factor by which the kinetic energy changes is 2.
So, when the runner doubles his speed, his kinetic energy will increase by a factor of 2.